What are all the possible rational zeros for y=x^4-x^3-4x^2+2x+4 and how do you find all zeros?

Apr 5, 2018

The zeros for this function are at $x = - \sqrt{2}$, $x = - 1$, $x = \sqrt{2}$, and , $x = 2$.

Explanation:

The possible rational zeros are the factors of the coefficients of the first and last term in standard form. The factors of 4 are $\pm 1$, $\pm 2$, and $\pm 4$. We note that when $x = 2$, $y = 0$. Therefore, $\left(x - 2\right)$ must be a factor of $y$. We can use synthetic division to prove this, but synthetic division is difficult to format properly in Socratic, so let's just write

${x}^{4} - {x}^{3} - 4 {x}^{2} + 2 x + 4$

$= {x}^{4} - 2 {x}^{3} + {x}^{3} - 2 {x}^{2} - 2 {x}^{2} + 4 x - 2 x + 4$

$= {x}^{3} \left(x - 2\right) + {x}^{2} \left(x - 2\right) - 2 x \left(x - 2\right) - 2 \left(x - 2\right)$

$= \left(x - 2\right) \left({x}^{3} + {x}^{2} - 2 x - 2\right)$

Now we note that $x = - 1$ is a zero of ${x}^{3} + {x}^{2} - 2 x - 2$ so $x + 1$ must be a factor of ${x}^{3} + {x}^{2} - 2 x - 2$.

$\left(x - 2\right) \left({x}^{3} + {x}^{2} - 2 x - 2\right)$

$= \left(x - 2\right) \left[{x}^{2} \left(x + 1\right) - 2 \left(x + 1\right)\right]$

$= \left(x - 2\right) \left(x + 1\right) \left({x}^{2} - 2\right)$

Now we can factor ${x}^{2} - 2$ as the difference of two squares.

$\left(x - 2\right) \left(x + 1\right) \left({x}^{2} - 2\right) = \left(x - 2\right) \left(x + 1\right) \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right)$

We can see that this expression evaluates to zero when $x = - \sqrt{2}$, $x = - 1$, $x = \sqrt{2}$, and , $x = 2$.